Question

A tank contain two layers of liquids, water (p= 1000 kg /m3) and mercury (p= 13600 kg / m3) . A cube with the edges of length 0.3 m and having the mass of 100 kg is lowered into the tank. Describe the position of cube when it reaches the equilibrium​

Answer 1

In equilibrium position 0.24 m of its height is in water while 0.6 m of its height is in mercury.

Explanation:

As shown in the figure let us assume that in equilibrium condition x m of the cube lies in water

Length of the cube in Mercury = 0.3-x

From the concept of buoyancy

Weight of the cube = Buoyant Force on the cube due to water + Buoyant force on the cube due to Mercury

$100g=0.3\times0.3\times x\times 1000\times g+(0.3-x)\times 0.3\times 0.3\times 13600\times g$

$\implies 100=0.3^2[1000x+(0.3-x)\times 13600]$

$\implies 100=0.09\times 100[10x+(0.3-x)\times 136]$

$\implies 100=9[10x+40.8-136x]$

$\implies 100=9(40.8-126x)$

$\implies 100=367.2-1134x$

$\implies 1134x=267.2$

$\implies x=\frac{267.2}{1134}$

$\implies x=0.24$ m

Therefore, length of the cube in mercury

$=0.3-0.24=0.06$m

Hence in equilibrium position 0.24 m of its height is in water while 0.6 m of its height is in mercury.

Hope this answer is helpful.

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